Download Solutions 2

MAT 425: Introduction to Analysis I
Solution #2
Set 4 → Rational Numbers. Prove the following statements (a, b ∈ Q).
(12) a) |a − b| ≤ |a| + |b|.
Note that | − a| = |a|. Then |a − b| = |a + (−b)| ≤ |a| + | − b| = |a| + |b|.
(12) b) |a| − |b| ≤ |a − b|.
It follows from a) that |a| = |b − (b − a)| ≤ |b| + |b − a|, so |a| − |b| ≤ |a − b|.
(12) c) |a| − |b| ≤ |a − b|.
Fact: u ≤ v and −u ≤ v implies |u| ≤ v. From b), we know that |a| − |b|
b|. If we
≤ |a −
interchange a and b, we have |b|−|a| ≤ |a−b|. Using Fact, this implies that |a|−|b| ≤ |a−b|.
Set 5 → Cauchy Sequences.
(14) a) Let {xk } and {yk } be two equivalent Cauchy sequences. Show that
equivalence of Cauchy sequences respects the properties of
Reflexivity: Show that {xk } ∼ {xk };
We want to show that beyond a certain m, we will have |xk − xk | < 1/n (∀n ∈ N and
for k ≥ m). But that will always be true since |xk − xk | = 0 for any k and 0 < 1/n.
Symmetry: Show that {xk } ∼ {yk } if {yk } ∼ {xk };
Same as above except that instead of |xk − xk | = 0 we use the fact that |xk − yk | =
|yk − xk |.
Transitivity: Show that {xk } ∼ {yk } and {yk } ∼ {zk } ⇒ {xk } ∼ {zk }.
See Lemma 2.1.1 in Book.
(10) b) What kinds of real numbers are representable by Cauchy sequences
of integers?
At some point in a Cauchy sequence (i.e. beyond a term xm , for some m), all terms
are going to be very close to each other (i.e. |xj − xk | < 1/n ∀n ∈ N, where j, k ≥ m,
and where 1/n can certainly be smaller than 1, which is the minimum distance between two
distinct integers). Since the terms in the sequence are integers, the only way the sequence
can be Cauchy is if, beyond the mth term, all terms are equal. For example, a Cauchy
sequence of integers could be (1, 3, 4, −50, 14, 13, −4, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, ...),
–2–
which means that only integers can be represented by Cauchy sequences of integers.
(+10) *c) Suppose {xk } = x1 , x2 , x3 , ... and {yk } = y1 , y2 , y3 , ... are two sequences
of rational numbers. Define the shuffled sequence to be x1 , y1 , x2 , y2 , x3 , y3 , ....
Prove that the shuffled sequence is Cauchy if and only if {xk } and {yk } are
equivalent Cauchy sequences.
Let {zk } = (x1 , y1 , x2 , y2 , ...).
(⇒) To show {zk } is Cauchy implies {xk } and {yk } are equivalent, we will show that
when {xk } and {yk } are not equivalent then {zk } is not Cauchy.
If {xk } and {yk } are not equivalent, then for any fixed m, we can find a k s.t. |xk −yk | >
1/n for some n. Knowing that zk = xk and zk+1 = yk , this means that |zk − zk+1 | is also
> 1/n, which means that {zk } is not Cauchy.
(⇐) Now, let’s assume that {xk } and {yk } are equivalent Cauchy sequences. This means
that ∀n ∈ N, ∃m s.t. |xk − yk | ≤ 1/2n for k ≥ m. Since both sequences are Cauchy, we can
say that |xj − xk | ≤ 1/2n and |yj − yk | ≤ 1/2n for j, k ≥ m.
Case 1: If zk = xk and zj = xj , then |zk − zj | = |xk − xj | ≤ 1/2n < 1/n.
Case 2: If zk = yk and zj = yj , then |zk − zj | = |yk − yj | ≤ 1/2n < 1/n.
Case 3: If zk = xk and zj = yj , then |zk − zj | = |xk − yj | = |xk + xj − xj − yj | ≤
|xk − xj | + |xj − yj | ≤ 1/2n + 1/2n = 1/n.
Case 4: If zk = yk and zj = xj , then |zk − zj | = |yk − xj | = |yk + yj − yj − xj | ≤
|yk − yj | + |yj − xj | ≤ 1/2n + 1/2n = 1/n.
(10) d) Can a Cauchy sequence of positive rational numbers be equivalent to
a Cauchy sequence of negative rational numbers?
Yes. Consider the sequence of positive rational numbers {1/k}. It is Cauchy since ∀n,
we can find an m s.t. |1/j − 1/k| ≤ 1/n for j, k ≥ m. We just need to take m ≥ 2n. We then
have |1/j − 1/k| ≤ |1/j| + |1/k| (from Set 4 a) ≤ 1/2n + 1/2n = 1/n. We can similarly show
that the sequence of negative rational numbers {−1/k} is also Cauchy. (Both sequences
represent the number 0). To show that the sequences are equivalent, we observe that for
any k, the distance between the terms of both sequences, |1/k − (−1/k)| = 2/k, will become
very small as k → ∞. Formally, ∀n, we can find m ≥ 2n s.t. |1/k − (−1/k)| = 2/k and
since k ≥ m and therefore, k ≥ 2n, we have 2/k ≤ 2/2n = 1/n.
(15) e) Show that the rationals are dense in R by showing that given any real
number x and error 1/n, there exists a rational number y such that |x − y| ≤ 1/n.
–3–
See Theorem 2.2.5 in Book.
(+10) *f ) Show that there are an infinite number of rational numbers in
between any two distinct real numbers.
The short solution is as follows: Consider the two real numbers to be x and y, where
x < y. Next consider the equivalence class of Cauchy sequences of rationals converging to
y. From this (infinite) class of Cauchy sequences, it is easy to find one that will be strictly
increasing (making all terms distinct) and such that the distance between all terms beyond
a certain point will be less than |y − x|.
The longer solution: First let’s show that between any two real numbers, there is a
rational number (which was In-Class Example #1, Set 3, b), i.e., show that ∀a, b ∈ R
(a < b), ∃q ∈ Q s.t. a < q < b.
Case 1: Suppose b − a > 1. Since the distance between a and b is greater than 1, we
can actually find an integer between a and b: if a ∈ Z, we then know that b > a + 1, so
take q = a + 1. If a ∈
/ Z, then take q = [a] + 1, where the [] means “integer part” (e.g.
[3.14159] = 3).
Case 2: Suppose b − a ≤ 1. Since b − a > 0 we know that (Axiom of Archimedes)
∃k ∈ N s.t. b − a > 1/k, which means that bk − ak > 1. So we can apply case 1 to ak and
bk instead of a and b and again, find an integer, say m s.t. ak < m < bk, which can be
rewritten a < m/k < b. Since m/k is rational, we then take q = m/k.
Having shown that there is a rational number between any two distinct real numbers,
we can certainly repeat the above procedure on a and q or on q and b. In fact, nothing
restricts us from repeating this infinitely.
Set 6 → Limits and Completeness.
(15) a) Let x1 , x2 , x3 , ... be a sequence of real numbers such that |xk | ≤ 1/2k ,
and set yk = x1 + x2 + ... + xk . Show that the sequence y1 , y2 , y3 , ... converges.
To show that the sequence converges, we can show that it is Cauchy. Fix 1/n. We need
to find a positive integer m s.t. ∀j, k ≥ m, we have |yj − yk | ≤ 1/n. We will use the fact that
P∞ 1
i=1 2i < ∞ (it’s a convergent series, which in fact is = 1). Without any loss of generality,
assume that j ≤ k.
j
k
k
k
k
X
X
X
X
X 1
1
1
|yj − yk | = xi −
xi = xi ≤
|xi | ≤
= j − k
i
2
2
2
i=1
i=1
i=j+1
i=j+1
i=j+1
And for any fixed 1/n, we can find a big enough m s.t. 21j − 21k < 1/n for any j, k ≥ m, which
shows that the sequence {yk } is Cauchy and thus converges. This argument is sufficient to
–4–
deserve full marks. However, if we want to be formal, we need to show what value of m
must be chosen. To find it, we can work backwards and find that m ≥ log2 n (with a little
algebra).
√
(+10) *b)
Prove that if 0 < a < 2, then a < 2a < 2. Prove that the sequence
q
p √
√ p √
2, 2 2, 2 2 2, ... converges. What is the limit? (Help: if limn→∞ an = l, then
√
√
limn→∞ 2an = 2l.)
√
If 0 < a < 2, then a2 < 2a < 4, so a < 2a < 2. This shows that
r q
q
√
√
√
2 < 2 2 < 2 2 2 < ... < 2,
so the sequence converges by Theorem 3.1.2 (Book). Denote the sequence
√ by {an }. Then
√
the sequence { 2an } is the same as {an+1 }. So the hint shows that l = 2l, or l = 2.